Problem: Find all real numbers $k$ for which there exists a nonzero, 2-dimensional vector $\mathbf{v}$ such that
\[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = k \mathbf{v}.\]Enter all the solutions, separated by commas.
Explanation: Let $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$.  Then
\[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = \begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 8y \\ 2x + y \end{pmatrix},\]and
\[k \mathbf{v} = k \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}.\]Thus, we want $k$, $x$, and $y$ to satisfy
\begin{align*}
x + 8y &= kx, \\
2x + y &= ky.
\end{align*}From the first equation, $(k - 1) x = 8y$.  If $x = 0$, then this equation implies $y = 0$.  But the vector $\mathbf{v}$ is nonzero, so $x$ is nonzero.  From the second equation, $2x = (k - 1) y$.  Similarly, if $y = 0$, then this equation implies $x = 0$, so $y$ is nonzero.  We also see that $k \neq 1$, because if $k = 1$, then $y = 0$, which again implies $x = 0$.

Hence, we can write
\[\frac{x}{y} = \frac{8}{k - 1} = \frac{k - 1}{2}.\]Cross-multiplying, we get $(k - 1)^2 = 16$.   Then $k - 1 = \pm 4.$  Therefore, $k = \boxed{5}$ or $k = \boxed{-3}$.

To make sure that these values of $k$ work, we should check if the corresponding vector $\mathbf{v}$ exists.  For $k = 5$, we can take $\mathbf{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, and for $k = -3$, we can take $\mathbf{v} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$, so both values of $k$ are possible.